Consider the following partial differential equation for with the constant c > 1:

Solution of this equation is \(\frac{{\partial u}}{{\partial y}} + c\frac{{\partial u}}{{\partial x}} = 0\)This question was previously asked in

PY 7: GATE ME 2017 Official Paper: Shift 1

Option 2 : u(x, y) = f(x – cy)

CT 1: Ratio and Proportion

2536

10 Questions
16 Marks
30 Mins

**Explanation:**

One approach is to put options and check if they satisfy the solution.

In this way putting option b

u (x,y) = f(x – cy)

\(\frac{{\partial u}}{{\partial y}} + C\frac{{\partial u}}{{\partial x}} = f'\left( {x - cy} \right)\left( { - c} \right) + cf'\left( {x - cy} \right) = 0\)

The solution of \(\frac{{\partial u}}{{\partial y}} + C\frac{{\partial u}}{{\partial x}} = 0\) is

U(x,y) = f (x – cy)

**Alternate solution**

Let u = f(ax + by)

\(\therefore \frac{{\partial u}}{{\partial \left( {ax + by} \right)}} = f'\left( {ax + by} \right)\)

Now \(\frac{{\partial u}}{{\partial y}} + C\frac{{\partial u}}{{\partial x}} = 0\)

\(\frac{{\partial u}}{{\partial \left( {ax + by} \right)}} \times \frac{{\partial \left( {ax + by} \right)}}{{\partial y}} + C\frac{{\partial u}}{{\partial \left( {ax + by} \right)}} \times \frac{{\partial \left( {ax + by} \right)}}{{\partial x}} = 0\)

⇒ b + ca = 0

b = - ac,

If a = 1

b = -c

∴ u = f(x - cy)